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In Exercises \(1-8,\) given \(y=f(u)\) and \(u=g(x),\) find \(d y / d x=\) \(f^{\prime}(g(x)) g^{\prime}(x)\) $$y=-\sec u, \quad u=\frac{1}{x}+7 x$$

Short Answer

Expert verified
\( \frac{dy}{dx} = (\sec(u) \tan(u)) (\frac{1}{x^2} - 7) \). Substitute \(u = \frac{1}{x} + 7x\) if necessary.

Step by step solution

01

Identify the given functions

We are given two functions: \(y = -\sec(u)\) and \(u = \frac{1}{x} + 7x\). Our task is to find \( \frac{dy}{dx} \).
02

Differentiate y with respect to u

To find \( \frac{dy}{dx} \), we first need \( \frac{dy}{du} \). Since \(y = -\sec(u)\), use the derivative of \(\sec(u)\) with respect to \(u\): \( \frac{d}{du}[ \sec(u) ] = \sec(u) \tan(u) \). Thus, \( \frac{dy}{du} = -( \sec(u) \tan(u) )\).
03

Differentiate u with respect to x

Next, we find the derivative of \(u\) with respect to \(x\): \(u = \frac{1}{x} + 7x\). Differentiate term by term: \( \frac{d}{dx} \left[ \frac{1}{x} \right] = -\frac{1}{x^2} \) and \( \frac{d}{dx}[7x] = 7 \). Thus, \( \frac{du}{dx} = -\frac{1}{x^2} + 7 \).
04

Apply the Chain Rule

Using the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \). Substitute the derivatives calculated: \( \frac{dy}{dx} = \left( -(\sec(u) \tan(u)) \right) \times \left( -\frac{1}{x^2} + 7 \right) \).
05

Simplify the Expression

Simplify \( \frac{dy}{dx} = (\sec(u) \tan(u)) \times (\frac{1}{x^2} - 7) \). Substitute \(u = \frac{1}{x} + 7x\) if further substitution is needed for specific evaluations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a fundamental concept in calculus. It deals with the idea of how a function changes as its input changes. In simpler terms, it's about finding the rate of change or the slope of a curve at any point. This rate of change is often referred to as the derivative. To perform differentiation, we apply specific rules or formulas that relate to the function's form.
  • For example, if we have a function like \(y = x^2\), the derivative \(\frac{dy}{dx}\) gives us \(2x\), which tells us how the output \(y\) changes as \(x\) changes.
  • The derivative of a linear function is constant, while for non-linear functions, it changes depending on \(x\).
When working with more complex functions, such as those involving \(\sec(u)\), we use specific differentiation formulas. The differentiation process becomes a combination of applying these rules, breaking down the function, and finding how each part contributes to the overall rate of change. This sets the stage for advanced techniques like the Chain Rule.
Composite Functions
Composite functions occur when we combine two or more functions into a single function. The result is a new function that relies on the transformations of the original functions. For example, if a function \(u = g(x)\) is used inside another function \(y = f(u)\), we get a composite function \(y = f(g(x))\).
  • This scenario often appears in calculus problems where the inner function \(g(x)\) modifies the input before it reaches the outer function \(f(u)\).
  • The input \(x\) is first transformed by \(g(x)\) before \(f\) takes over, which affects the output \(y\).
In our given example, the function \(u = \frac{1}{x} + 7x\) is plugged into \(y = -\sec(u)\), effectively forming the composite \(y = -\sec\left(\frac{1}{x} + 7x\right)\). Understanding how functions interact to form composite functions is crucial. This understanding helps greatly when applying the Chain Rule, which allows us to find derivatives of such functions.
Secant Function
The secant function, denoted as \(\sec(u)\), is a trigonometric function. It is the reciprocal of the cosine function, meaning \(\sec(u) = \frac{1}{\cos(u)}\). Here are some key points about the secant function:
  • It is undefined wherever \(\cos(u) = 0\), leading to vertical asymptotes in its graph.
  • Like other trigonometric functions, it repeats its pattern in a periodic way.
In calculus, when differentiating secant, the derivative is given by \(\frac{d}{du}[ \sec(u) ] = \sec(u) \tan(u)\). This formula is crucial because it tells us how \(\sec(u)\) changes as \(u\) changes. In our exercise, this derivative was used to find \(\frac{dy}{du}\), which is part of solving the derivative using the Chain Rule. Understanding the behavior of \(\sec(u)\) and its interactions with other functions helps when working through calculus problems involving trigonometric expressions.

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Most popular questions from this chapter

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